Download INTEGRATION TECHNIQUES Math 101 Contents 1. Substitution

INTEGRATION TECHNIQUES
Math 101
Contents
1. Substitution (§§4.3, 4.8)
2. Integration
by parts
R
R (§7.2)
R
2.1. R xn sin x dx, R xn cos x dx, xn ex dx
2.2. R ex sin x dx, ex cos x dx
2.3. xn (ln x)m dx
3. Rational functions (§7.3) [advanced]
3.1. Step 1. Reducing an improper fraction
3.2. Step 2. Decomposing into partial fractions
3.3. Step 3. Integrating partial fractions
4. Trigonometric functions (§7.4, including exercises)
4.1. The universal trigonometric
substitution [advanced]R
R
4.2. Integrals of the form R R(sin2 x, cos x) sin x dx and R(sin x, cos2 x) cos x dx
4.3. Integrals of
R the form R(sin x, cos x) dx
4.4. Integrals R tann x dx
R
R
4.5. Integrals sin mx sin nx dx, sin mx cos nx dx, cos mx cos nx dx
5. Exponential and hyperbolic functions
6. Trigonometric and hyperbolic substitutions (§7.4)
6.1. Example
7. Other irrationalities
8. Table integrals
1
2
2
2
2
2
2
2
2
3
3
3
3
3
3
4
4
4
4
5
References: Thomas & Finney, Calculus and Analytic Geometry, 9th Edition:
4.3, 4.8, 7.2, 7.3 [advanced], 7.4.
There is no general rule for evaluating an integral. Each time you have to look for a formula/rule that would
simplify the integral and eventually reduce it to table ones.
1. Substitution (§§4.3, 4.8)
Use these formulas (direct and backward):
Z
u = g(x)
f (g(x))g 0 (x) dx =
= f (u) du = F (u) + C = F (g(x)) + C,
du = g 0 (x) dx
Z
Z
x = g(u)
f (x) dx =
= f (g(u))g 0 (u) du = F (u) + C = F (g −1 (x)) + C,
dx = g 0 (u) du
Z
Z
1
f (x) dx = F (x) + C (important special case).
f (kx + b) dx = F (kx + b) + C, where
k
Z
For definite integrals the former formula takes the form
Z
a
b
Z g(b)
g(b)
u = g(x)
f (g(x))g (x) dx =
=
f (u) du = F (u)
.
0
du = g (x) dx
g(a)
g(a)
0
R
Do not forget that dx also needs substitution! (In an expression like f (g(x)) dx you cannot substitute
u = g(x) !) Do not forget to return to the original variable (or recalculate limits in case of definite integrals)!
Some standard substitutions will be discussed later (4, 5, 6, 7).
1
2
MATH 101
2. Integration by parts (§7.2)
Z
Z
u dv = uv −
Z
v du,
a
b
b Z
u dv = uv −
a
b
v du.
a
Success of the method depends on the decomposition ofR the integrand in the form u dv. Choose for dv
something
R that you can easily integrate (as you need u = du for the formula)! Proceed only if the resulting
integral
v du is simpler that the original one and you know what to do with it! (Certain integrals, e.g.,
R x
e sin x dx below, are an exception: integration by parts may give the same integral, but the resulting
expression can be treated as an equation.) Below are some typical examples.
R
R
R
2.1. xn sin x dx, xn cos x dx, xn ex dx. Take u = xn . Integration by parts is to be applied n times (with
u = xn ), each time reducing the power of x by 1.
R
R
2.2. ex sin x dx, ex cos x dx. Integrate by parts twice, each time taking u = ex . Solve the resulting
equation in the original integral.
R
2.3. xn (ln x)m dx. Take u = (ln x)m . Integration by parts reduces m by 1. Do it m times.
3. Rational functions (§7.3)
[advanced]
A rational function P (x)/Q(x), where P and Q are polynomials in x, can be integrated in three steps:
(1) Reduce the fraction if it is improper (i.e., deg P > deg Q);
(2) Decompose P (x)/Q(x) into a sum of partial fractions;
(3) Integrate each partial fraction.
3.1. Step 1. Reducing an improper fraction. Do not overlook this step! Otherwise, Step 2 will
produce an erroneous result! If the fraction is improper, i.e., deg P > deg Q, then divide P by Q and write
P/Q = F + (R/Q), where F is the ratio and R is the remainder. The polynomial F (x) can easily be
integrated. With the proper fraction R(x)/Q(x) proceed to Step 2.
3.2. Step 2. Decomposing into partial fractions. From now on assume that the fraction P (x)/Q(x)
is proper (i.e., deg P < deg Q, see above). We need to decompose the denominator Q(x) into the product
of linear (i.e., of the form (x − r) ) and irreducible quadratic (i.e., of the form (x2 + px + q) with p2 − 4q <
0) factors. Now the decomposition into partial fractions can be found by the method of undetermined
coefficients. Write the identity P (x)/Q(x) = . . . , where the right hand side is composed as follows: each
linear factor (x − r)m contributes m terms
A1
A2
Am
+
+ ... +
,
2
x − r (x − r)
(x − r)m
and each quadratic factor (x2 + px + q)n contributes n terms
B1 x + C1
B2 x + C2
Bn x + Cn
+
+ ... + 2
.
x2 + px + q (x2 + px + q)2
(x + px + q)n
(Of course, each coefficient in the resulting expression should be given it’s own name. The total number of
undetermined coefficients A, B, C, . . . must be equal to the degree of Q.) In the resulting identity get rid of
the denominators (by multiplying both the sides by Q(x) ) and write down equations for the undetermined
coefficients. There are two ways to obtain equations: (1) by equating the coefficients of equal powers of x
(typically, simpler equations are obtained from higher powers), and (2) by substituting particular values
of x (usually some simple values like x = 0, ±1, etc. or the values x = r, where (x − r) is a factor in the
decomposition of Q(x) ).
Important Remark: The resulting system of linear equations in A, B, C, · · · must have a unique
solution! If there is no solutions, something is wrong in your calculation. If there are many solutions,
then either something is wrong or there is not enough equations.
3.3. Step 3. Integrating partial fractions. Here are the formulas (with C omitted):
Z
Z
dx
dx
1
= ln|x − r|,
=
(substitution x − r = t).
x−r
(x − r)n
(1 − n)(x − r)n−1
For the other partial fractions, first complete the square:
Z
Z
Bx + C
Bt + C 0
p 2
4q − p2
Bp
dx
=
dt
(where
t
=
x
+
,
a
=
> 0, and C 0 = C −
).
2
n
2
2
n
(x + px + q)
(t + a )
2
4
2
INTEGRATION TECHNIQUES
3
Then
Z
1
t dt
= ln(t2 + a2 ),
t2 + a2
2
Z
1
t dt
=
(t2 + a2 )n
2(1 − n)(t2 + a2 )n−1
(substitution t2 + a2 = u),
Z
dt
1
t
= tan−1 is a table integral, and the remaining integral In =
t2 + a2
a
a
steps via integration by parts and reducing n:
In =
1
a2
Z
dt
is evaluated in n
(t2 + a2 )n
Z
t2 + a2 − t2
1
1
1
dt
=
I
−
t
·
d
n−1
(t2 + a2 )n
a2
2(n − 1)a2
(t2 + a2 )n−1
1
t
1
1
= [by parts] = 2 In−1 −
+
In−1 .
2
2
2
n−1
a
2(n − 1)a (t + a )
2(n − 1)a2
Z
4. Trigonometric functions (§7.4, including exercises)
An integral of a rational function of sin x and cos x can always be reduced to integrating a rational
function. The best thing to try is using trigonometric identities (see transc.pdf) to convert products to
sums and to reduce powers. Below are a few standard hints.
4.1. The universal trigonometric substitution [advanced]. One lets t = tan(x/2). Then x = 2 tan−1 t
and
2t
1 − t2
2dt
,
cos
x
=
,
dx =
sin x =
1 + t2
1 + t2
1 + t2
are expressed in terms of t rationally.
Important Remark: This substitution is universal. However, typically it leads to a huge amount of
calculations. Thus, use it only if you cannot think of a better approach!
Important Remark: Always try to simplify the expression using trigonometric identities. The general rule is the following: the bigger the argument of the functions (sin and cos), the lower the degree
of the expression. However, keep in mind that in general you want to have all functions of the same
argument! Below are a few examples.
R
R
4.2. Integrals of the form R(sin2 x, cos x) sin x dx and R(sin x, cos2 x) cos x dx, where R is a rational
function. Use the substitution cos x = t, sin2 x = 1 − t2 , sin x dx = −dt (in the former case) or sin x = t,
cos2 x = 1 − t2 , cos x dx = dt (in the latter case).
Example.
Z
Z
3
Z
2
sin x(1 − sin x) cos x dx =
sin x cos x dx =
(u − u3 )du =
sin2 x sin4 x
−
+C
2
4
(where u = sin x and du = cos x dx).
R
4.3. Integrals of the form R(sin x, cos x) dx, where R is a rational function whose all terms have even
degree In other words, the integrand can be expressed in terms of sin2 x, cos2 x, and sin x cos x. Reduce the
degree using the formulas
sin2 x =
1 − cos 2x
,
2
cos2 x =
1 + cos 2x
,
2
sin x cos x =
1
sin 2x (see transc.pdf).
2
Example.
Z
sin x cos3 x dx =
4.4. Integrals
R
Z
1
1 + cos 2x
sin 2x
dx =
2
2
Z sin 2x sin 4x
+
4
8
dx = −
cos 2x cos 4x
−
+ C.
8
16
tann x dx. One has
Z
Z
tan x dx = − ln|cos x|,
n
tan x dx =
Z
n−2
tan
x 1−
1 dx =
cos2 x
Z
tann−2 x dx −
tann−1 x
.
n−1
R
R
R
4.5. Integrals sin mx sin nx dx, sin mx cos nx dx, cos mx cos nx dx. Convert products of functions to
sums (see trigonometric identities in transc.pdf).
4
MATH 101
5. Exponential and hyperbolic functions
R
An integral of the form R(ex ) dx (where R is a rational function) can be reduced
to integrating a rational
R
function by the substitution ex = t, x = ln t, dx = dt/t. An integral of the form R(sinh x, cosh x) dx can be
treated either by expressing sinh x and cosh x in terms of ex or, better yet, similar to 4, using corresponding
hyperbolic identities (see transc.pdf).
6. Trigonometric and hyperbolic substitutions (§7.4)
Assume that we need to integrate an expression rational in x and the radical
complete the square
D
b 2
− 2 ,
ax2 + bx + c = a x +
2a
4a
√
ax2 + bx + c. First,
where, as usual, D = b2 − 4ac,
b
and make the substitution u = x + 2a
, du = dx. Now, depending on the signs of a and D, we need to treat
p
√
√
√
one of the following three irrationalities: u2 + r2 , u2 − r2 , or r2 − u2 (where r = |D/4a2 |).
6.1. Example.
√
3 + 6x − 3x2 =
p
p
√ √
√
−3(x2 − 2x − 1) = −3((x − 1)2 − 2) = 3 2 − u2 . Here r = 2.
Now we make an appropriate trigonometric or hyperbolic substitution and reduce the given integral to a
trigonometric or hyperbolic one, which can be treated as in 4 or 5, respectively.
Irrationality
√
√
√
r2
−
u2
Trigonometric substitution
u = r sin t, −π/2
√ 6 t 6 π/2
du = r cos t dt,
r2 − u2 = r cos t
−1
t = sin (u/r)
Hyperbolic substitution
−
u2 + r 2
u = r tan t, −π/2√
< t < π/2
du = r dt/ cos2 t,
u2 + r2 = r/ cos t
−1
t = tan (u/r)
u = r sinh t, t any
√ number
du = r cosh t dt,
u2 + r2 = r cosh t
−1
t = sinh (u/r)
u2 − r 2
u = r/ cos t, 0 6 t < π/2
√ or π/2 < t 6 π
du = r sin t dt/ cos2 t,
u2 − r2 = ∓r tan t
−1
t = cos (r/u)
u = ±r cosh t, t > √
0
du = ±r sinh t dt,
u2 − r2 = r sinh t
−1
t = cosh (±u/r)
Important Remark: To my opinion, in the last two cases the hyperbolic substitutions are usually
simpler.
Important Remark: In the last case one should be very careful about the signs. The integrand is
defined on two disjoint intervals, x > 1 and x 6 −1, which should be treated differently. (The upper
sign corresponds
to the first interval.) One should keep in mind that cosh t > 1 for any t and that, by
√
definition, u2 − r2 > 0 for any u.
7. Other irrationalities
r
ax + b
ax + b
An expression rational in x and m
is integrated by the substitution
= tm . Then x =
cx + d
cx + d
dtm − b
− m
and the integrand becomes rational (see 3).
ct − a
INTEGRATION TECHNIQUES
5
8. Table integrals
Z
Z
xn dx =
xn+1
+C
n+1
Z
(n 6= 1),
ex dx = ex + C,
Z
sin x dx = − cos x + C,
Z
sec2 x dx = tan x + C,
Z
sec x tan x dx = sec x + C,
Z
sinh x dx = cosh x + C,
Z
dx
x
= sin−1 + C (a > 0),
2
a
−x
Z
x
dx
1
√
= sec−1 + C (a > 0),
2
2
a
a
x x −a
Z
dx
−1 x
√
= sinh
+ C (a > 0),
a
x2 + a2
Z
1
x
dx
= tanh−1 + C (a > 0).
a2 − x2
a
a
√
a2
dx
= ln x + C,
x
Z
ax
+ C,
ax dx =
ln a
Z
cos x dx = sin x + C,
Z
csc2 x dx = − cot x + C,
Z
csc x cot x dx = − csc x + C,
Z
cosh x dx = sinh x + C,
Z
dx
1
x
= tan−1 + C (a > 0),
2
2
a +x
a
a
Z
√
dx
x
= cosh−1 + C
2
a
−a
x2
(see transc.pdf for the expressions for sinh−1 , cosh−1 , and tanh−1 in terms of ln.)
[advanced]
This topic has been omitted or moved to Math 102
(a > 0),